Question

Assertion:\[P{{F}_{5}}\], \[S{{F}_{6}}\] and \[{{H}_{2}}S{{O}_{4}}\] are the examples of expanded octet molecules.
Reason: Octet rule is not applicable to second period elements of the periodic table
A.Both assertion and reason are correct and reason is the correct explanation for assertion.
B.Both assertion and reason are correct but reason is not the correct explanation for assertion.
C.Assertion is correct but reason is incorrect
D.Both assertion and reason are incorrect.

Answer 1

Hint: To select the correct option, we should know about expanded octet. We should know about exceptions about expanded octet molecules.

Step by step answer:
So, first of all we'll know about expanded octet. The important point that we should focus on is that the main group elements that form more bonds than would be predicted by the octet rule are called hypervalent compounds, and have what we call that they have expanded octet. It means that there are more than eight electrons around one atom.
We should note that the octet rule can be expanded by some elements by using the d-orbitals found in the third principal energy level and beyond. The common examples of elements that form an expanded octet are Sulphur, phosphorus, silicon, and chlorine.
So, we should know about hypervalent compounds. We should know that hypervalent molecules are molecules that contain one or more main group elements that have more than eight electrons in their valence levels as a result of bonding. Phosphorus pentafluoride (\[P{{F}_{5}}\]), sulfur hexafluoride (\[S{{F}_{6}}\]), are examples of hypervalent molecules.
Let us know about phosphorus pentafluoride. We should know that the central atom phosphorus does not follow octet rule. And this element is of third period phosphorus pentafluoride. In the \[P{{F}_{5}}\]molecule, the central phosphorus atom is bonded to five F atoms, thus having 10 bonding electrons and violating the octet rule.
Now, let us take\[S{{F}_{6}}\]. We should know that in the \[S{{F}_{6}}\] molecule, the central sulphur atom is bonded to six fluorine atoms, so sulphur has 12 bonding electrons around it. It is also a third period element.
Now, we will take example\[{{H}_{2}}S{{O}_{4}}\]. In sulphuric acid (\[{{H}_{2}}S{{O}_{4}}\]), each oxygen has a full octet (eight valence electrons), whereas sulphur has an expanded octet (twelve valence electrons). And sulphur as we know is from period 2.
So, from the above discussion we can now say that assertion is correct. Reason is incorrect. So, option C is the correct answer.


Note:
Let us know about free radicals. They also un follow octet rule. We should know that there are a few stable molecules which contain an odd number of electrons. We call these molecules, by the name of free radicals. It contains at least one unpaired electron, which we can call a clear violation of the octet rule. Let us know about examples of free radicals and they are nitrogen (II) oxide, nitrogen (IV) oxide, and chlorine dioxide.